# Misc 15 - Chapter 6 Class 12 Application of Derivatives

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Misc 15 Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is 4 3 . Given, Radius of sphere = r Let R be the radius of the cone and H be its height. Let BOC = Now, AC = AO + OC H = r + r cos H = r (1 + cos ) And, R = r sin We need to maximize volume of cone. Volume of the cone is V = 1 3 2 V = 1 3 2 sin 2 r (1 + cos ) V = 1 3 3 sin 2 (1 + cos ) Finding = 1 3 3 2 sin cos (1+ cos )+ sin 2 ( sin ) = 1 3 3 2 sin cos (1+ cos ) sin 3 = 1 3 3 2 sin cos (1+ cos ) sin 3 = 1 3 3 sin 2 cos + 2 cos 2 sin 2 = 1 3 3 sin 2 cos + 2 cos 2 (1 cos 2 ) = 1 3 3 sin 3 cos 2 + 2 1 = 1 3 3 sin 3 cos 2 + 3 cos 1 = 1 3 3 sin 3 cos +1 1 cos +1 = 1 3 3 sin 3 1 +1 Putting = 0 1 3 3 sin +1 3 1 = 0 sin (cos + 1) (3 cos 1) = 0 So, = 1 3 Thus, H = r (1 + cos ) H = r 1 + 1 3 H = 4 3 This value is either maxima or minima. So, finding 2 2 = 1 3 3 sin 3 cos2 + 2 cos 1 = 1 3 3 cos 3 cos 2 +2 cos 1 + sin 6 cos 2 sin = 1 3 3 3 cos 3 +2 cos 2 cos + sin 6 cos sin 2 sin = 1 3 3 3 cos 2 +2 cos 2 cos 6 sin 2 cos 2 sin 2 Now, cos = 1 3 And sin2 = 1 cos2 = 1 1 3 2 = 1 1 9 = 8 9 Putting values in 2 2 2 2 = 1 3 3 3 1 3 3 +2 1 3 2 1 3 6 1 3 8 9 2 8 9 = 1 3 3 3 1 27 +2 1 9 1 3 6 8 27 2 8 9 = 1 3 3 1 9 + 2 9 1 3 16 9 16 9 = 1 3 3 32 9 = 32 3 27 Thus, 2 2 < 0 H = Hence proved

Chapter 6 Class 12 Application of Derivatives

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.